Complex Multiplications
Given an arithmetic of
dual numbers x+ey, and a conjugation # that turns e into –e, then consider this
operation:
x*y = ( x*y + (#x)*y + x*(#y) – (#x)*(#y) ) / 2

where * is
dualarithmetic multiplication: e*e = 1.
Then these laws follow:
*
is commutative
*
distributes over addition
If r is real, then x*(r*y) = r *(x*y)
1*1 = 1
e*e = 1
From which follows:
*
is isomorphic to complex multiplication.
Here are three complex
multiplications of triple ratios:
x*_{1}y
= ( x*y +_{1} (1#x)*y +_{1}
x*(1#y) –_{1} (1#x)*(1#y) ) / 2_{1}
x*_{2}y
= ( x*y +_{2} (2#x)*y +_{2}
x*(2#y) –_{2} (2#x)*(2#y) ) / 2_{2}
x*_{3}y
= ( x*y +_{3} (3#x)*y +_{3}
x*(3#y) –_{3} (3#x)*(3#y) ) / 2_{3}

where 2_{1} =
(1;2;2) ; 2_{2}
= (2;1;2) ; 2_{3}
= (2;2;1)
Exercise for the ambitious student:
Find formulas for x*_{1}y in terms of x_{1},
x_{2}, x_{3}, y_{1}, y_{2}, and y_{3};
and likewise for x*_{2}y and x*_{3}y.
Open Questions
How do
the three arithmetics relate, other than by S_{3} conjugation? Are
there identities involving products of additions, reductions, or complex
multiplications?
Define,
for any triple ratios A = (a_{1}, a_{2}, a_{3}) and B = (b_{1}, b_{2}, b_{3})
, the two functions M and m:
M(A,B) = (A+_{1}B)*(A+_{2}B)*(A+_{3}B)
/ (A*B)
= 1
/ ( a_{2}b_{3} +
a_{3}b_{2} ; a_{1}b_{3} + a_{3}b_{1}
; a_{1}b_{2} + a_{2}b_{1}
)
m(A,B) = (A[+]_{1}B)*(A[+]_{2}B)*(A[+]_{3}B)
/ (A*B)
= (a_{1}b_{1}(a_{2}b_{3}+a_{3}b_{2})
; a_{2}b_{2}(a_{1}b_{3}+a_{3}b_{1})
; a_{3}b_{3}(a_{1}b_{2}+a_{2}b_{1}))
Since
(x[+]_{a}y)*(x+_{a}y) =
x*y, it follows that:
M(A,B)*m(a,b) = A*B
m(a,b) = A*B / M(A,B)
1/m(A,B) = M(1/A, 1/B)
1/M(A,B) = m(1/A, 1/B)
For any triple ratios A,
B and C:
M(A,A) = A
m(A,A) = A
A * M(B,C) = M(A*B, A*C)
A * m(B,C) = m(A*B, A*C)
A / M(B,C) = m(A/B, A/C)
A / m(B,C) = M(A/B, A/C)
And for
any index a:
a#
M(A,B) = M(a#A, a#B)
a#
m(A,B) = m(a#A, a#B)
M and m are symmetric
under S_{3}, like times and divide.
How do M and m relate to
the three additions?
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